10. A parallel-plate capacitor has square plates of side s 2.50cm and +0 plate s
ID: 1779980 • Letter: 1
Question
10. A parallel-plate capacitor has square plates of side s 2.50cm and +0 plate separation d = 2.50mm. The capacitor is charged by a battery to a charge Q=4.00AC, after which the battery is disconnected. A porcelain dielectric (K = 6.5) is then inserted a distance y = 1.00cm into the capacitor as shown in the figure. Hint: Consider the system as two capacitors connected in parallel. Dielectric a) What is the effective capacitance of this capacitor? b) How much work is done in inserting the dielectric?Explanation / Answer
before inseting dielectric
capacitance of the capacitor co = eo*A/d = eo*s^2/d = 8.84*10^-12*0.025^2/(2.5*10^-3)
co = 2.21*10^-12 F
A = area of plate = s^2
after the dielctric is inserted
capacitance of upper part c1 = eo*s*(s-y)/d
capacitance of lower part c2 = k*eo*s*y/d
C1 and C2 are parallel
Ceff = C1 + C2 = (eo*s/d)*( s - y + ky)
Ceff = (8.84*10^-12*0.025/(2.5*10^-3))*(0.0025 - 0.01 + (6.5*0.01))
Ceff = 7.07*10^-12 F
===============
b)
initial energy Ui = (1/2)*Q^2/Co = (1/2)*(4*10^-6)^2/(2.21*10^-12) = 3.62 J
final energy Uf = (1/2)*Q^2/Ceff = (1/2)*(4*10^-6)^2/(7.07*10^-12) = 1.31 J
work = dU = 2.31 J
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